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Common formulations of the general measurement postulate of quantum mechanics have two parts, one establishing the probability of the outcome $m$ corresponding to an observable $M_m$ for a system in an initial state $\left|\psi\right>$:

$$\left|\left|\psi_m\right>\right|^2=p(m) $$

where $\left|\psi_m\right>\equiv M_m\left|\psi\right>$; and a second part specifying the post measurement state of the system given that $m$ is observed.

But is the seond part necessary? Isn't it a consequence of the first part?

The first part establishes a relationship between component norms and probabilities of the form

$$\Vert\left| \psi\right>\Vert_E=\sqrt{\mathbb{P}(E)}$$

where $\Vert\left| \psi\right>\Vert_E$ is the norm of the component of a system $\left| \psi\right>$ corresponding the the occurrence of event $E$, specifically, the postulate states that

$$\Vert\left| \psi\right>\Vert_{O=m\wedge M=M_m}=\sqrt{\mathbb{P}(O=m\wedge M=\psi_m)}$$

where $O$ is the observed outcome and $M$ is the applied observable.

Is this correspondence between norms and probability a general one so that

$$\Vert\left| \psi\right>\Vert_{M=M_m\vert O=m}=\frac{\Vert\left| \psi\right>\Vert_{O=m\wedge M=M_m}}{\sqrt{\mathbb{P}(O=m)}}=\frac{\Vert\left| \psi_m\right>\Vert}{\sqrt{p(m)}}$$

and the second part of the postulate thus follows from the first part (keeping in mind that, by construction, the event $O=m'\wedge M=M_{m''}$ cannot occur for $m'\neq m''$)?

If so, isn't the second part of the postulate entirely unnecessary: establishing this "conditional amplitude" is its only content?

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  • $\begingroup$ This seems like a way of formulating the two postulates as one. I know them as "We get an eigenvalue as the measurement value", and "The wavefunction collapses to the projection of the original wavefunction on the eigenspace of the measured eigenvalue". You instead formulate it as "We get an eigenvalue as the measurement value and the wavefunction collapses". $\endgroup$ – Omry Oct 11 '17 at 16:57
  • $\begingroup$ @Omry: There's no "collapse". That can be dispensed with (that's part of the point here). And your formulation mentions nothing about probabilities, which is the core of the the Born rule (which cannot be dispensed with). $\endgroup$ – orome Oct 11 '17 at 17:14
  • $\begingroup$ A minor detail: your expression for $p(m)$ is off. If you want the probability of measurement, you need to project your state onto the eigenstate corresponding to $m$, and then norm-square. $\endgroup$ – CDCM Oct 11 '17 at 17:34
  • $\begingroup$ @CDCM: I think you've misread: $\left|\psi_m\right>\equiv M_m\left|\psi\right>$, as stated in the question. $\endgroup$ – orome Oct 11 '17 at 17:58
  • $\begingroup$ Perhaps, though that equation isn't exactly clear what you're saying. e.g. I'd be more used to $$p(m)=\lvert\langle \phi_m \lvert \psi \rangle\rvert^2,$$ where $\phi_m$ are defined through the eigenvalue equation for the particular operator, $$\hat{M}\phi_m = \lambda_m \phi_m.$$ $\endgroup$ – CDCM Oct 11 '17 at 18:39

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